stringMatch("xxcaazz", "xxbaaz") → 3
stringMatch("abc", "abc") → 2
stringMatch("abc", "axc") → 0

import doctest


def all_two_chars_occurencies(string):
    """
    >>> list(all_two_chars_occurencies('abcd'))
    ['ab', 'bc', 'cd']
    >>> list(all_two_chars_occurencies('xxcaazz'))
    ['xx', 'xc', 'ca', 'aa', 'az', 'zz']
    """
    for index, char in enumerate(string[:-1]):
        yield char + string[index + 1]


def common_two_chars_occurences(a, b):
    """
    Given 2 strings, a and b, return the number of the positions where
    they contain the same length 2 substring.

    >>> common_two_chars_occurences('xxcaazz', 'xxbaaz')
    3
    >>> common_two_chars_occurences('abc', 'abc')
    2
    >>> common_two_chars_occurences('abc', 'axc')
    0
    """
    equal_duets = 0
    for a_duet, b_duet in zip(all_two_chars_occurencies(a),
                              all_two_chars_occurencies(b)):
        if a_duet == b_duet:
            equal_duets += 1
    return equal_duets


if __name__ == "__main__":
    doctest.testmod()